Q.Find the principal value of the following: $\tan^{-1}(-\sqrt{3})$

The key idea is that the principal value branch of $\tan^{-1}$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$, and we need the angle whose tangent is $-\sqrt{3}$.

Step 1: Recall that $\tan(\frac{\pi}{3}) = \sqrt{3}$.

Step 2: Since $\tan$ is an odd function, $\tan(-\frac{\pi}{3}) = -\sqrt{3}$. …

Why Inverse Trigonometric Graphs Matter

When you see $\tan^{-1}(x)$, you're not just "undoing" tan — you're picking a specific angle from a restricted domain. The tangent function repeats every $\pi$, so there are infinitely many angles whose tangent is $-\sqrt{3}$. For example, $\tan(\frac{2\pi}{3}) = -\sqrt{3}$, and $\tan(-\frac{\pi}{3}) = -\sqrt{3}$, and so on.

The "principal value" is the unique angle that lies in the principal branch of the inverse tangent function. For $\tan^{-1}$, this branch is defined as:

This interval is open — it excludes $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ because tangent is undefined there. So we must find an angle strictly between $-90^\circ$ and $90^\circ$ (in radians, between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$) whose tangent equals $-\sqrt{3}$.

Step-by-step solution

1. Recall the standard angle.

   We know $\tan(\frac{\pi}{3}) = \sqrt{3}$. Since tangent is an odd function ($\tan(-\theta) = -\tan\theta$), it follows that $\tan(-\frac{\pi}{3}) = -\sqrt{3}$.

2. Check the range.

   The angle $-\frac{\pi}{3}$ is approximately $-60^\circ$, which lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$. So it is a valid candidate for the principal value.

3. Confirm uniqueness. …

Method: Principal value of an inverse tangent

Use this whenever you must evaluate $\tan^{-1}$ of a specific number and report a single angle.

Steps

Step 1: Write down the principal branch

The inverse tangent always returns an angle in the open interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Your final answer must lie strictly inside this interval.

Step 2: Find the reference angle from a standard value

Ignore the sign of the argument for a moment and recall which standard angle produces that tangent magnitude (e.g. $\tan\frac{\pi}{3} = \sqrt{3}$, $\tan\frac{\pi}{4} = 1$, $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$).

Step 3: Restore the sign using oddness …

Common Mistakes

Mistake 1: Choosing $\frac{2\pi}{3}$ because its tangent is also $-\sqrt{3}$

Why it's wrong: $\frac{2\pi}{3}$ lies outside the principal branch $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, so it cannot be a principal value even though its tangent matches. Correct approach: pick the angle with tangent $-\sqrt{3}$ that lies inside the branch, namely $-\frac{\pi}{3}$. …